Cusp-end in the universal covering

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Let $M$ be a n-dimensional hyperbolic manifold with finite volume. Then as a consequence of the Margulis-Lemma we have a decomposition in different types of ends.

So let $C$ be a cusp-end. Then there is a set $C' \subset C$ s.t. $C'$ is isometric to $V \times [0, \infty)$ and $V$ is compact. So we can write a point in $C'$ as $(x,t)$.

Let $H^n$ be the universal cover, looking at it in the upper-half-space model, $R ^{n-1} \times R^+$ with pointsof the form $(x_1, ...,x_n)$. Let $\infty$ be the point corresponding to the cusp-end $C$.

Well, this is all known.

My question: Is there a possibility to make a connection between $t$ and $x_n$? So that I can say, if the point $(x_0, t_0)$ is fixed, then I know, that in the universal covering $x_n$ is at least ... ?

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One can choose a locally isometric universal covering map $F : \mathbb{H}^n \mapsto M$ so that the restriction of $F$ to $\mathbb{R}^{n-1} \times [1,+\infty)$ is a universal covering map over $C' \approx V \times [0,\infty)$. In more detail, you can choose a universal covering map $f : \mathbb{R}^{n-1} \mapsto V$ so that if $(x_1,…,x_{n-1},x_n) \in \mathbb{R}^{n-1} \times [1,+\infty)$ then $$F(x_1,…,x_{n-1},x_n) = (f(x_1,…,x_{n-1}),x_n-1) $$ which gives the correspondence $x_n-1 \leftrightarrow t$.